Question

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ᵠᵘᵉˢᵗⁱᵒⁿ-

A light particle moving horizontally with a speed 12 m/s strikes a very heavy block moving in the same direction at 10m/s. The collision is one dimensional and elastic. After the collision the particle will
ᗩ) move at 12m/s opposite to its original direction
ᗷ) move at 8m/s in its original direction ᑕ) move at 8m/s opposite to its original
ᗞ) move at 2m/s in its original direction

‼Nᴏᴛᴇ: ᴘʀᴏᴠɪᴅᴇ ᴛʜᴇ ᴀɴsᴡᴇʀ ᴡɪᴛʜ sᴏʟᴜᴛɪᴏɴ‼

⚠️___No spams____⚠️​

Answer 1

$\huge\underline\bold\red{AnswEr}$

Conservation of momentum,

$m_1v_1 + m_ 2v_2 =m_1v'_1 + m_2v'_2 \\ \\ \implies \: m_ 1(v_ 1 - v'_ 1) = m_ 2(v_2 - v'_2).......(1) \\ \\ conservation \: \: of \:kinetic \\ energy \: \: requires \\ \\ \frac{1}{2} (m_1 {v}^{2} _1 + m_ 2 {v}^{2} _2) = \frac{1}{2} (m_1 {v'}^{2} _1 + m_ 2 {v'}^{2} _2) \\ \\ \implies \: m_ 1( {v}^{2} - {v'}^{2} _1)= m_ 2( {v}^{2} _2 - {v'}^{2} _2) \\ \\ \implies \: m_ 1(v_ 1 - v'_ 1)(v_ 1 + v'_ 1) = m_ 2(v_ 2 - v'_ 2)(v_ 2 + v'_ 2) .......(2) \\ \\ dividing \: (2) \: and \: (1) \\ \\ v_ 1 - v'_ 1 = {v}^{2} _ 2 - v'_ 2 \\ \\ 12 + v'_ 1 = 10 + 10 \\ \\ v'_ 1 = 8m/s$

Answer 2

Given:

Particle :

mass = m₁

velocity v₁ = 12m/s

heavy block:

mass = M

velocity V₁= 10m/s

To Find:

Find the velocity of the particle 1 after the collision, v₂

Solution:

As the block is heavy than the particle so its velocity after collision will remain unchanged. V₂=V₁=10m/s

As the collision is elastic, e=1

∴ e = velocity of separation/ velocity of approach

      = (V₂-v₂)/(V₁-v₁)

1=(10-v₂)/(10-12)

2=10-v₂

v₂=10-2

   =8m/s

Hence the velocity of particle after collision is 8m/s.